The three stresses being equal,
Compressive strain = /c (a + ft)
Tensile strain = /t (a + /3)
.*. Shear strain = 2/s(a
shear stress i
shear strain 2 (a + /3)
Again, if the one inch cube be compressed equally on all
Compressive strain = jfa - 2//3
Volumetric decrease = 3 (fa - 2//3) = $f(a - 2/3)
unit stress = i
.-. K -
strain per inch cube 3 (a - 2/3)
Collecting the results, we have
E = ~
3 (a-2)3) 3(M-2)
a t5K + 2C
and the moduli are connected in value, if M be known.
Pp. 425 and 462. Strength and Stiffness of Shafts.
Example yr.—A steel marine engine weigh shaft has the dimensions
shewn in Fig. 917, where A and B are the bearings, and the levers are
as indicated. It is required to find the diameter of shaft in order that
tfoe stress shall not exceed 9000 Ibs. per sq. in., the angle of torsion
being limited to one degree in 20 ft. of length. (Actual case.)
T J 'I
Load on reversing lever
and neglecting Bra, which is not considerable, greatest Tm is at centre,
caused by the pull on I. P. and L.P. drag rods.
Tm « (4.500+5000) 18 = 171,000 Ib.ins.